To determine if the complex number [tex]\(-1 + i\sqrt{3}\)[/tex] is equal to [tex]\(2 \text{ cis } 120^\circ\)[/tex] in polar form, we need to verify both the magnitude (or modulus) and the angle (or argument) of the complex number in polar form.
Step 1: Compute the Magnitude
The magnitude [tex]\(r\)[/tex] of a complex number [tex]\(a + bi\)[/tex] is given by the formula:
[tex]\[ r = \sqrt{a^2 + b^2} \][/tex]
For the complex number [tex]\(-1 + i\sqrt{3}\)[/tex]:
- The real part [tex]\(a\)[/tex] is [tex]\(-1\)[/tex].
- The imaginary part [tex]\(b\)[/tex] is [tex]\(\sqrt{3}\)[/tex].
Let's compute the magnitude:
[tex]\[ r = \sqrt{(-1)^2 + (\sqrt{3})^2} \][/tex]
[tex]\[ r = \sqrt{1 + 3} \][/tex]
[tex]\[ r = \sqrt{4} \][/tex]
[tex]\[ r = 2 \][/tex]
So, the magnitude of [tex]\(-1 + i\sqrt{3}\)[/tex] is [tex]\(2\)[/tex].
Step 2: Compute the Angle
The angle [tex]\(\theta\)[/tex] of a complex number [tex]\(a + bi\)[/tex] in polar form is given by:
[tex]\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \][/tex]
Note: The angle should be adjusted based on the quadrant in which the complex number lies.
For the complex number [tex]\(-1 + i\sqrt{3}\)[/tex]:
[tex]\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right) \][/tex]
This evaluation gives an angle in the second quadrant because the real part is negative and the imaginary part is positive. The exact angle is:
[tex]\[ \theta = 180^\circ - \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) \][/tex]
[tex]\[ \theta = 180^\circ - 60^\circ \][/tex]
[tex]\[ \theta = 120^\circ \][/tex]
So, the angle of [tex]\(-1 + i\sqrt{3}\)[/tex] in polar form is [tex]\(120^\circ\)[/tex].
Conclusion
- The magnitude of [tex]\(-1 + i\sqrt{3}\)[/tex] is [tex]\(2\)[/tex].
- The angle of [tex]\(-1 + i\sqrt{3}\)[/tex] is [tex]\(120^\circ\)[/tex].
Therefore, the complex number [tex]\(-1 + i\sqrt{3}\)[/tex] indeed corresponds to [tex]\(2 \text{ cis } 120^\circ\)[/tex] in polar form.
The statement [tex]\(-1 + i\sqrt{3} = 2 \text{ cis } 120^\circ\)[/tex] is True.
